Complex numbers and polynomials - University College Cork

We can also conclude that the graph of p is symmetric about the line x = − band that it has a single turning point at (− b , 4ac−b2 ). Moreover, this is a minimum2a 4apoint if a > 0, and a maximum point if a < 0, i.e.,{ ≥4ac−b 2, if a > 0,p(x)4a≤ 4ac−b2 , if a < 0,4awith equality here iff x = − b . In other words,2a2a ,if a > 0, andif a < 0.min{p(x) : x ∈ R} =max{p(x) : x ∈ R} =4ac − b24a4ac − b24aExample 1. If a, b, c are real numbers, and a ≠ 0, thenp(x) = ax 2 + bx + c ≥ 0, ∀x ∈ R,iff a > 0, c ≥ 0 and b 2 ≤ 4ac. These conditions hold iff p is the square of the modulusof a linear polynomial.Proof. Suppose p ≥ 0 on R. Then, in particular, c = p(0) ≥ 0. Next, for x ≠ 0,a + b x + c x 2 = p(x) 1 x 2 ≥ 0,and so, letting x → ∞, we deduce that a ≥ 0. But, a ≠ 0. Hence, a > 0, so that4ac − b 24a= p(− b2a ) ≥ 0,whence b 2 ≤ 4ac. Thus the stated conditions are necessary to ensure the nonnegativityof p. Conversely, if they hold, then, by completing the square, we see thatp(x) = a(x + b 4ac − b22a )2 +4a≥4ac − b24a≥ 0, ∀x ∈ R,from which it also follows that p is the square of the modulus of the linear polynomialwhich has complex coefficients.□√ ax +b + √ 4ac − b 2 i2 √ a9